Mathematics Test-3

Let, \(y=x^{\sin x}\)

taking \(\log\) both side we get,

\(\log y=\log x^{\sin x}\)

differentiate both side with respect to \(x\).

\(\frac{d}{d x} \log y=\frac{d}{d x} \log x^{\sin x}\)

\(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x} \sin x \log x\)

\(\frac{1}{y} \frac{d y}{d x}=\sin x \frac{d}{d x} \log x+\log x \frac{d}{d x} \sin x\)

\(\frac{1}{y} \frac{d y}{d x}=\sin x \times \frac{1}{x}+\log x \cos x\)

\(\frac{d y}{d x}=y \times\left(\frac{\sin x}{x}+\log x \cos x\right)\)

\(\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x+x \cos x \log x}{x}\right)\)

Given:

\(A=\) The set of lines which are parallel to the \(x\)-axis

and, \(B=\) The set of numbers which are multiples of 5

As we know that, we can draw infinite number of lines parallel to the \(x\)-axis.

So, \(A\) is an infinite set.

Similarly, there are \(n\) numbers that can be a multiple of 5 i.e.,

\(B=\{5 n: n \in Z\}\)

We can see that, \(B\) is also an infinite set.

Therefore,both \(A\) and \(B\) are infinite sets.

Let the given statement be \(P(n)\), i.e.,

\({P}(n):(2 n+7)<(n+3)^{2}\)

Put \(n=1\) in \(P (n)\)

\(2\times 1+7=9<(1+3)^{2}=16\), which is true.

Let \(P (k)\) be true for some positive integer \(k\),

\((2 k+7)<(k+3)^{2} \ldots(1)\)

We shall now prove that \(P (k+1)\) is true whenever \(P (k)\) is true.

\(\Rightarrow\{2(k+1)+7\}=(2 k+7)+2 <(k+1+3)^{2}\)

\(\Rightarrow\{2(k+1)+7\}=(2 k+7)+2 <(k+4)^{2}\)

\(\Rightarrow\{2(k+1)+7\}=(2 k+7)+2 <(k^2+16+8k)\)

\(\Rightarrow\{2(k+1)+7\}=(2 k+9) <(k^2+16+8k)\)

\(\Rightarrow\{2(k+1)+7\}=(2 k+9) <(k^2+9+7+6k+2k)\)

\(\Rightarrow\{2(k+1)+7\}=(2 k+9) <((k+3)^{2}+2k+7).......(2)\)

From \((1)\) and \((2)\), it is clear that, \(P(k+1)\) is true whenever \(P(k)\) is true.

So, by the principle of mathematical induction, statement \(P(n)\) is true for all natural numbers.

We have \((x+y)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} . y+{ }^{n} C_{2} x^{n-2}. y^{2}+\ldots+{ }^{n} C_{n} y^{n}\)

General term: General term in the expansion of \((x+y)^{n}\) is given by:

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{n} C_{\mathrm{r}} \times x^{\mathrm{n}-\mathrm{r}} \times y^{\mathrm{r}}\)

We have to find term independent of \(x\) in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{10}\).

We know that,

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times {x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow \mathrm{T}_{(\mathrm{r}+1)}={ }^{10} \mathrm{C}_{\mathrm{r}} \times\left({x}^{2}\right)^{10-\mathrm{r}} \times\left(\frac{-1}{{x}^{3}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-2 \mathrm{r}} \times\left({x}^{3}\right)^{-\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-2 \mathrm{r}} \times({x})^{-3 \mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-5 \mathrm{r}}\)

For the term independent of \(x\), power of \(x\) should be zero

Therefore, \(20-5 r=0\)

\(\Rightarrow \mathrm{r}=4\)

\(\mathrm{T}_{(4+1)}=(-1)^{4} \times{ }^{10} \mathrm{C}_{4}={ }^{10} \mathrm{C}_{4}\)

As we know,

Standard equation of a hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\).

Coordinates of foci are \((\pm a e, 0)\).

Eccentricity \(e=\sqrt{1+\frac{b^{2}}{a^{2}}}\)

Given,

Coordinates of foci \(=(\pm 3,0)\) and eccentricity, \(e=\frac{3}{2}\)

So, \(ae =3\)

As \(e=\frac{3}{2}\)

\(\therefore a\times \frac{3}{2} =3\)

\(\Rightarrow a=\frac{3\times2}{3}\)

\(\Rightarrow a=2\)

Eccentricity \(e=\sqrt{1+\frac{b^{2}}{a^{2}}}\)

Squaring both sides, we get

\(e^2=1+\frac{b^{2}}{a^{2}}\)

\(\Rightarrow e^2-1=\frac{b^{2}}{a^{2}}\)

\(\therefore b^{2}=a^{2}\left(e^{2}-1\right)\)

Putting the value of \(a\) and \(e\), we get

\( b^{2}=2^{2}\left[\left(\frac{3}{2}\right)^{2}-1\right]\)

\(\Rightarrow b^{2}=4 \left(\frac{9}{4}-1\right)\)

\(\Rightarrow b^{2}=4\left(\frac{9-4}{4}\right)\)

\(\Rightarrow b^{2}=4 \times \frac{5}{4}\)

\(\Rightarrow b^{2}=5\)

So, the equation of the required hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{5}=1\).

The modulus value is not negative.

\(\tan ^{-1}(-x)=-\tan ^{-1}(x)\)

Given, equation is \(x^{2}+5|x|-6=0\)

\(\Rightarrow\left|x^{2}\right|+5|x|-6=0\)

\(\Rightarrow\left|x^{2}\right|+6|x|-|x|-6=0\)

\(\Rightarrow|x|(|x|+6)-1(|x|+6)=0\)

\(\Rightarrow(|x|+6)(|x|-1)=0\)

\(\Rightarrow(|x|+6)=0\) and \((|x|-1)=0\)

\(\Rightarrow|{x}|=-6\) and \(|{x}|=1\)

But \(|x|=-6\) which is not possible because value of modulus is not negative.

\(\Rightarrow|x|=1\)

\(\Rightarrow x=1\) and \(x=-1\)

Given, \(\alpha\) and \(\beta\) are toots of the equation \(x^{2}+5|x|-6=0\)

Thus, \(\alpha=1\) and \(\beta=-1\).

Now, consider, \(\mid\tan ^{-1} \alpha-\tan ^{-1} \beta|=| \tan ^{-1}(1)-\tan ^{-1}(-1) \mid\)

\(=\left|\tan ^{-1}(1)+\tan ^{-1}(1)\right|\)

\(=\left|2 \tan ^{-1}(1)\right|\)

\(=2. \frac{\pi}{4}\)

\(=\frac{\pi}{2}\)

The word ALLAHABAD contains 9 letters, in which A occur 4 times, L occurs twice and the rest of the letters occur only once.

Number of Permutations of ‘\(n\)’ objects where there are \(n_1\) repeated items, \(n_2\) repeated items, \(n_k\) repeated items taken ‘\(r\)’ at a time:

\(p(n, r)=\frac{n !}{n_{1} ! n_{2} ! n_{3} ! \ldots n_{k} !}\) 

Therefore,  Number of different words formed by the word ALLAHABAD using all the letters \(=\frac{9 !}{4 ! \times 2 !}\)

\(=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 2}\) \(=7560\)

Now, let us take both L together and consider (LL) as 1 letter.

Then, we will have to arrange 8 letters, in which A occurs 4 times and the rest of the letters occur only once.

So, the number of words having both \(L\) together will be \(=\frac{8 !}{4 !}\)

\(=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\) \(=1680\)

Therefore, the number of words with both L not occurring together will be \(= 7560 - 1680\) \(= 5880\)

Let us consider two AP's having first term \(a _{1}\) and \(a _{2}\) and the common difference are \(d _{1}\) and \(d _{2}\) respectively.

According to the question,

\(\frac{ S _{1}}{ S _{2}}=\frac{3 n +8}{7 n +15} \)

As we know,

\(S_n=\frac{n}{2}[2a+(n-1) \times d]\)

\(\therefore \frac{\left.\frac{ n }{2}\left[2 a _{1}+( n -1) d _{1}\right)\right]}{\left.\frac{ n }{2}\left[2 a _{2}+( n -1) d _{2}\right)\right]}=\frac{3 n +8}{7 n +15} \)

\(\Rightarrow \frac{\left.\left[ a _{1}+\left(\frac{ n -1}{2}\right) d _{1}\right)\right]}{\left[ a _{2}+\left(\frac{ n -1}{2}\right) d _{2}\right]}=\frac{3 n +8}{7 n +15} \)...(1)

As we know,

\(n^{\text{th}}\) term of AP \((a_n)=a+(n-1) \times d\)

So the ratio of \(12^{\text {th }}\) term of first AP and second AP \(=\frac{ a _{1}+(12-1) d _{1}}{ a _{2}+(12-1) d _{2}}\)

\(=\frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}\)...(2)

From (1) and (2), we can write

\(\frac{ n -1}{2}=11\)

\(\Rightarrow n-1=22\)

\(\Rightarrow n=23\)

Putting \(n = 23\) in equation (1), we get

\(\frac{\left.\left[ a _{1}+\left(\frac{23 -1}{2}\right) d _{1}\right)\right]}{\left[ a _{2}+\left(\frac{ 23 -1}{2}\right) d _{2}\right]}=\frac{3 \times 23 +8}{7 \times 23 +15} \)

\(\Rightarrow \frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}=\frac{69+8}{161+15}\)

\(\Rightarrow \frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}=\frac{77}{176}\)

\(\Rightarrow \frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}=\frac{7}{16}\)

\(\Rightarrow \frac{12^{\text {th }} \text { term of first AP}}{12^{\text {th }} \text {term of second AP}}=\frac{7}{16}\)

So, the ratio of \(12^{\text {th }}\) term offirst AP and second APis \(7:16\).

Given:

\(\lim _{\mathrm{x} \rightarrow 3} \frac{\mathrm{x}^{4}-81}{\mathrm{x}^{3}-27}\)

On checking the limit by putting \(x=3\) in above equation, we get \((\frac{0}{0})\) form.

We know that:

L-Hospital Rule as:

\(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

\(\therefore\) By using the L-Hospital rule, we get,

\(\lim _{x \rightarrow 3} \frac{x^{4}-81}{x^{3}-27}=\lim _{x \rightarrow 3} \frac{\frac{d}{dx}(x^{4}-81)}{\frac{d}{dx}(x^{3}-27)}\)

\(=\lim _{x \rightarrow 3} \frac{4 x^{3}}{3 x^{2}}\quad\) \((\because \frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.)\)

\(=\lim _{x \rightarrow 3} \frac{4 x}{3}\)

Putting the value of \(x \rightarrow3\) in above equation, we get

\(=\frac{4(3)}{3}\)

\(=4\)

Vector equation of a plane whose perpendicular distance from the origin is \(d\) and \(\hat{n}\) is the unit normal vector to the plane through origin is given by \(\vec{r} . \hat{n}=d\)

Now,

Let \(\vec{n}=\hat{i}+2 \hat{j}+2 \hat{k}\) is the normal to the required plane from the origin and it is at a distance of \(\frac{1}{3}\) units from the origin.

\(\Rightarrow|\vec{n}|=|\hat{i}+2 \hat{j}+2 \hat{k}|=\sqrt{1^2+2^2+2^2}=3 \text {. }\)

So, the unit normal vector \(\hat{n}=\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\)

As we know that, the equation of a plane whose perpendicular distance from the origin is \(d\) and \(\hat{n}\) is the unit normal vector to the plane through origin is given by \(\vec{r} . \hat{n}=d\)

Here, \(d =\frac{1 }{ 3}, \hat{n}=\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\) and let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)

\(\Rightarrow \frac{1}{3} x+\frac{2}{3} y+\frac{2}{3} z=\frac{1}{3} \)

\(\Rightarrow x +2 y +2 z =1 \)

So, the equation of the required plane is \(x+2 y+2 z=1\)

\(f^{\prime}(x)=\left(|x|-x^{2}\right) e^{-x^{2}}+\left(|x|-x^{2}\right) e^{-x^{2}}, x \geq 0\)

\(f^{\prime}=2\left(x-x^{2}\right) e^{-x^{2}}\)

hence option (D) is wrong

\(g^{\prime}(x)=x e^{-x^{2}} 2 x \)

\(f^{\prime}(x)+g^{\prime}(x)=2 \times e^{-x^{2}} \)

\(f(x)+g(x)=-e^{-x^{2}}+c \)

\(f(x)+g(x)=-e^{-x^{2}}+1 \)

\(F(\ell n 3)+g(\sqrt{\ell n 3})=1-\frac{1}{3}=\frac{2}{3} \text { (option (A) is wrong) }\)

\(\mathrm{H}(\mathrm{x})=\psi_{1}(\mathrm{x})-1-\alpha \mathrm{x}=\mathrm{e}^{-\mathrm{x}}+\mathrm{x}-1-\alpha \mathrm{x}, \qquad \mathrm{x} \geq 1 \& \alpha \in(1, \mathrm{x})\)

\(\mathrm{H}(1)=\mathrm{e}^{-1}+1-1-\alpha<0\)

\(\mathrm{H}^{\prime}(\mathrm{x})=-\mathrm{e}^{-\mathrm{x}}+1-\alpha>0\)

\( \Rightarrow \mathrm{H}(\mathrm{x})\) is \(\downarrow\)

\(\Rightarrow\) option \((\mathrm{B})\) is wrong

(C) \(\psi_{2}(\mathrm{x})=2\left(\psi_{1}(\beta)-1\right)\)

Applying L.M.V.T to \(\psi_{2}(\mathrm{x})\) in \([0, \mathrm{x}]\)

\(\psi_{2}^{\prime}(\beta)=\frac{\psi_{2}(x)-\psi_{2}(0)}{x} \)

\(2 \beta-2+2 \mathrm{e}^{-\beta}=\frac{\psi_{2}(\mathrm{x})-0}{\mathrm{x}}\)

\(\Rightarrow \psi_{2}(\mathrm{x})=2 \mathrm{x}\left(\psi_{1}(\beta-1)\right)\) has one solution

Given,

The curve \(y=a \sqrt{x}+b x\) passes through the point \((1,2)\)

Thus, the point \((1,2)\) will satisfy the curve \(y=a \sqrt{x}+b x\).

\(\Rightarrow 2=a(1)+b(1)\)

\(\Rightarrow a+b=2\)

Also, given the area bounded by the curve, line \(x=4\) and \(x\)-axis is \(8\) sq.unit

\(\text { Area }=\int_{0}^{4}(a \sqrt{x}+b x) d x\)

\(\Rightarrow 8=\left[a \frac{x^{\frac{3}{2}}}{\frac{3}{2}}+b \frac{x^{2}}{2}\right]_{0}^{4}\)

\(\Rightarrow 8=\frac{2 a}{3} \cdot 8+8 b\)

\(\Rightarrow 2 a+3 b=3 \ldots(2)\)

Solve equation (1) and (2) simultaneously, we get

\(a=3\) and \(b=-1\)

Given,

\(\frac{p^{2}}{a^{2}}+k \cos \alpha+\frac{q^{2}}{b^{2}}=\sin ^{2} \alpha......\)(i)

\(\cos ^{-1}\left(\frac{p}{a}\right)+\cos ^{-1}\left(\frac{q}{b}\right)=\alpha\)

As we know,

\(\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}\right)\)

\(\cos ^{-1}\left(\frac{p q}{a b}-\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)=\alpha\)

\(\cos \alpha=\left(\frac{p q}{a b}-\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\right)\)

\(\frac{p q}{a b}-\cos \alpha=\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}}\)

Squaring both sides, we get

\(( \frac{p q}{a b}-\cos \alpha)^2=(\sqrt{1-\frac{p^{2}}{a^{2}}} \sqrt{1-\frac{q^{2}}{b^{2}}} )^2\)

\(\frac{(p q)^{2}}{(a b)^{2}}+\cos ^{2} \alpha-2 \frac{p q}{a b} \cos \alpha=\left(1-\frac{p^{2}}{a^{2}}\right)\left(1-\frac{q^{2}}{b^{2}}\right)\)

\(\frac{(p q)^{2}}{(a b)^{2}}+\cos ^{2} \alpha-2 \frac{p q}{a b} \cos \alpha=1-\frac{p^{2}}{a^{2}}-\frac{q^{2}}{b^{2}}+\frac{(p q)^{2}}{(a b)^{2}}\)

\(\sin ^{2} \alpha=\frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}-2 \frac{p q}{a b} \cos \alpha.....\)(ii)

Comparing equation (i) and (ii), we get

\(\mathrm{k}=-\frac{2 p q}{a b}\)

Given, \(\tan ^{2} \theta=1-e^{2}\)

\(\therefore \sec \theta+\tan ^{3} \theta \cdot \operatorname{cosec} \theta\)

\(=\sec \theta+\tan ^{2} \theta \cdot \tan \theta \cdot \operatorname{cosec} \theta\)

\(= \sec \theta+\tan ^{2} \theta \cdot \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta}\)

\(= \sec \theta+\tan ^{2} \theta \cdot \sec \theta\)

\(= \sec \theta\left(1+\tan ^{2} \theta\right)\)

\(= \sqrt{1+\tan ^{2} \theta} \cdot\left(1+\tan ^{2} \theta\right)\)

\(=\left(1+\tan ^{2} \theta\right)^{\frac{3}{2}}\)

\(=\left(1+1-e^{2}\right)^{\frac{3}{2}}\)

\(=\left(2-e^{2}\right)^{\frac{3}{2}}\)

The dot product of two vectors:

Let and be two vectors then if the angle between them is then we write the dot product of the two vectors as:

\(\vec{\mathrm{a}}.\vec{\mathrm{b}}=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta\)

It is given that \(\vec{\mathrm{a}}+\vec{\mathrm{b}}+\vec{\mathrm{c}}=0\)

Therefore, \((\vec{\mathrm{a}}+\vec{\mathrm{b}})=-\vec{\mathrm{c}}\).

Squaring on both sides we get,

\((\vec{\mathrm{a}}+\vec{\mathrm{b}})^{2}=(-\vec{\mathrm{c}})^{2}\)

\(|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}+2 \vec{\mathrm{a}} .\vec{\mathrm{b}}=|\vec{\mathrm{c}}|^{2}\)

\(|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}+2|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta=|\vec{\mathrm{c}}|^{2}\)

\((4)^{2}+\left(3^{2}\right)+2(4)(3) \cos \theta=(\sqrt{37})^{2}\)

\(16+9+24 \cos \theta=37\)

\(\cos \theta=\frac{1}{2}\)

\(\theta=60^{\circ}\)